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MSU Graduate Theses
Spring 2016
On The Number Of Distinct Cyclic Subgroups Of
A Given Finite Group
Joseph Dillstrom
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Dillstrom, Joseph, "On The Number Of Distinct Cyclic Subgroups Of A Given Finite Group" (2016). MSU Graduate Theses. Paper
2547.
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ON THE NUMBER OF DISTINCT CYCLIC SUBGROUPS OF A
GIVEN FINITE GROUP
A Masters Thesis
Presented to
The Graduate College of
Missouri State University
In Partial Fulfillment
Of the Requirements for the Degree
Master of Science, Mathematics
By
Joseph A. Dillstrom
May 2016
ON THE NUMBER OF DISTINCT CYCLIC SUBGROUPS OF A GIVEN
FINITE GROUP
Mathematics
Missouri State University, May 2016
Master of Science
Joseph A. Dillstrom
ABSTRACT
In the study of finite groups, it is a natural question to consider the number of distinct cyclic subgroups of a given finite group. Following an article by M. Tărnăuceanu
in the American Mathematical Monthly, we consider arithmetic relations between
the order of a finite group and the number of its cyclic subgroups. We classify several infinite families of finite groups in this fashion and expand upon an open problem posed in the article.
KEYWORDS: abstract algebra, group theory, finite group, cyclic subgroup, Lagrange’s Theorem,
This abstract is approved as to form and content
Dr. Richard Belshoff
Chairperson, Advisory Committee
Missouri State University
ii
ON THE NUMBER OF DISTINCT CYCLIC SUBGROUPS OF A
GIVEN FINITE GROUP
By
Joseph A. Dillstrom
A Masters Thesis
Submitted to The Graduate College
Of Missouri State University
In Partial Fulfillment of the Requirements
For the Degree of Master of Science, Mathematics
May 2016
Approved:
Dr. Richard Belshoff, Chairperson
Dr. Les Reid, Member
Dr. Cameron Wickham, Member
Dr. Julie J. Masterson, Graduate College Dean
iii
ACKNOWLEDGEMENTS
I would first like to thank Dr. Joseph Siler at OTC, who showed me the
beauty of mathematics in a way that no one had before, and inspired me to pursue
the subject as a field of study.
Second, I would like to thank my thesis advisor, Dr. Richard Belshoff, for
his patience and for his kindness. The combination of his rigorous approach to the
subject and his easygoing style of teaching were a perfect fit for the way I learn and
he has been a very positive infuence. I would also like to thank Dr. Les Reid for his
interest in the project and his feedback throughout.
Finally, I would like to thank my parents, whose affection and support mean
the world to me.
iv
TABLE OF CONTENTS
1.
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2.
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
3.
LOWER END RESULTS FOR |C(G)| . . . . . . . . . . . . . . . . .
5
4.
HIGHER END RESULTS FOR |C(G)|. . . . . . . . . . . . . . . . . .
17
5.
CONCLUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
v
LIST OF TABLES
Table 1. Groups of Small Order (Up To Isomorphism)...............................................4
Table 2. Calculations and Observations of |C(G)|.......................................................5
vi
1. INTRODUCTION
In the study of finite groups, it is natural to consider their cyclic subgroup
structure. Since every element generates a finite cyclic subgroup, determining the
number of distinct cyclic subgroups of a given finite group G can give a sense of
how many “transformations” of elements are possible within the group. In the American Mathematical Monthly article which served as the starting point for our research, M. Tărnăuceanu notes that this problem is relatively unexplored in the literature [1]. For groups of small order this can be accomplished relatively quickly by
hand, but applying the same procedure to groups of large order can be computationally unfeasible.
In this spirit we characterize certain arithmetic relations between the order of a finite group |G| and the number of its distinct cyclic subgroups |C(G)|.
We also classify the cyclic subgroup structure of certain infinite families of finite
groups. In particular we characterize the finite groups satisfying |C(G)| = k for
k ∈ {1, . . . , 4}. We also supply alternative proofs of the theorems in the AMM article that characterize the finite groups satisfying |C(G)| = |G| and |C(G)| = |G| − 1.
We conclude by expanding the open problem posed by Tărnăuceanu to characterize
the finite groups satisfying |C(G)| = |G| − 2.
1
2. BACKGROUND
In this section we will fix our notation and briefly review some properties
of finite groups. Given a finite group G, we will use the notation C(G) for the set
of its cyclic subgroups, and we will use |C(G)| for the cardinality of C(G). We will
denote a generic cyclic group of order n by Cn .
We begin by stating some results of cyclic groups that we will make use of
throughout the paper. Since cyclic groups themselves are well understood, we state
the results without proof. The statements of the results are taken from Dummit
and Foote’s Abstract Algebra [2].
PROPOSITION 2.1: Let G = hxi. If |x| = n < ∞, then |xa | =
n
,
(n,a)
where
(n, a) = gcd(n, a).
PROPOSITION 2.2: Assume that H = hxi is cyclic and |x| = n < ∞. Then
H = hxa i if and only if (a, n) = 1. In particular, the number of generators of H is
φ(n), where φ(n) is Euler’s phi-function.
PROPOSITION 2.3: Let H be a cyclic group. If |H| = n < ∞, then for each
positive integer a dividing n there is a unique cyclic subgroup of H of order a. Furthermore, for every integer m, hxm i = hx(n,m) i, so that the subgroups of H correspond bijectively with the positive divisors of n. Thus, |C(H)| = τ (n), where τ (n)
is the number of divisors function.
Two cornerstones of finite group theory are Lagrange’s Theorem and Cauchy’s
Theorem. We make extensive use of both in our results and state them here without proof.
THEOREM 2.4: (Lagrange’s Theorem) If G is a finite group and H is a subgroup
of G, then the order of H divides the order of G, i.e. |H| | |G|.
Two useful corollaries of Lagrange’s Theorem are as follows:
2
COROLLARY 2.5: If G is a finite group and x ∈ G, then the order of x divides
the order of G.
COROLLARY 2.6: If G is a group of prime order p, then G is cyclic, hence
G∼
= Cp .
By Proposition 2.3, the converse of Lagrange’s Theorem holds in a cyclic
group. Thus, for a finite group G, if H ≤ G is a cyclic subgroup of order m with
divisors d1 , d2 , . . . , dk , then H (hence G) contains cyclic subgroups Ci of orders di ,
i ∈ {1, . . . , k}. We will occasionally make use of this observation when we need to
establish a contradiction, so we record it here as a corollary.
COROLLARY 2.7: Let G be a finite group, H be a cyclic subgroup of G of order
m, and d1 , . . . , dk be the positive divisors of m. Then G contains cyclic subgroups
Cdi , i ∈ {1, . . . , k}.
THEOREM 2.8: (Cauchy’s Theorem) If G is a finite group and p is a prime dividing |G|, then G contains an element of order p.
The advantage of Cauchy’s Theorem in our research is that we can guarantee a lower bound of |C(G)| by considering the prime factorization of |G|. In certain instances this will allow us to narrow down the number of distinct prime powers of |G| subject to certain assumptions about |C(G)|.
We also make use of the first part of Sylow’s Theorem throughout our paper.
The notation and statement of the theorem also come from Dummit and Foote [2].
DEFINITION 2.9: If G is a group of order pα m, where p - m, then a subgroup of
order pα is called a Sylow p-subgroup of G. The number of Sylow p-subgroups of G
will be denoted np .
THEOREM 2.10: (Sylow 1) Let G be a group of order pα m, p - m. Then Sylow
p-subgroups exist, i.e. np ≥ 1.
We will also rely on certain results concerning semi-direct products and nor-
3
mal subgroups, and we will introduce these results as they become necessary.
4
3. LOWER END RESULTS FOR |C(G)|
We begin by considering the groups of order less than or equal to 10 (up
to isomorphism). Since there are relatively few of these groups, we use them as a
“baseline” for considering the subgroup structure of higher order groups in related
families.
Table 1: Groups of small order (up to isomorphism)
|G| # Of Groups
Isomorphism Type
1
1
hei
2
1
C2
3
1
C3
4
2
C4 , C2 × C2 ∼
= V4
5
1
C5
6
2
C6 , D6 ∼
= S3
7
1
C7
8
5
C8 , C4 × C2 , C23 , D8 , Q8
9
2
C9 , C3 × C3
10
2
C10 , D10
Observe that nearly all groups (up to isomorphism) of order less than or
equal to 10 are either cyclic, dihedral, or direct products of cyclic groups. The one
outlier is the quaternion group Q8 . Since |C(G)| for cyclic groups G is given by
Proposition 2.3, we seek characterizations of |C(G)| for direct products of cyclic
groups and for the dihedral group D2n . Our main objective is to characterize the
finite groups satisfying |C(G)| = k for k ∈ {1, . . . , 4}. This will tell us which finite groups have relatively few cyclic subgroups and provide “lower end” results for
|C(G)|.
5
By hand calculations we record |C(G)| for these groups of small order and
note the arithmetic relations between |G| and |C(G)| that arise in Table 2.
Table 2: Calculations and observations of |C(G)|
G
|C(G)|
Note
hei
1
|C(G)| = 1
C2
2
|C(G)| = |G|
C3
2
|C(G)| = |G| − 1
C4
3
|C(G)| = |G| − 1
C2 × C2
4
|C(G)| = |G|
C5
2
|C(G)| = 2
C6
4
|C(G)| = |G| − 2
D6 ∼
= S3
5
|C(G)| = |G| − 1
C7
2
C8
4
C4 × C2
6
|C(G)| = 2
|G|
|C(G)| =
2
|C(G)| = |G| − 2
C23
8
|C(G)| = |G|
D8
7
|C(G)| = |G| − 1
Q8
5
C9
3
C3 × C3
5
|C(G)| = |G| − 3
|G|
|C(G)| =
3
|C(G)| = |G| − 4
C10
4
|C(G)| = |G| − 6
D10
7
|C(G)| = |G| − 3
The groups of small order all seem to have similar arithmetic relations between
their orders and their number of cyclic subgroups. In fact, we can easily dispense
of the case when |C(G)| = 1.
6
PROPOSITION 3.1: Let G be a finite group. Then |C(G)| = 1 if and only if
G = hei.
Proof. The reverse implication follows immediately by Lagrange’s Theorem. For
the converse, suppose that |C(G)| = 1. Then G must be the trivial group, for if G
contained another element then it would generate a cyclic subgroup distinct from
hei, contrary to hypothesis.
From the table it appears that the only finite groups G satisfying |C(G)| = 2
are the cyclic groups of prime order. This is indeed the case.
PROPOSITION 3.2: Let G be a finite group. Then |C(G)| = 2 if and only if
G = Cp for some prime p.
Proof. (⇐) Let G = Cp for some prime p. Then G is cyclic, so that every subgroup
of G is cyclic, and by Lagrange’s Theorem every cyclic group of order p has exactly
two subgroups, namely hei and G.
(⇒) Suppose that |C(G)| = 2 and write |G| = pα1 1 pα2 2 · · · pαk k . Certainly hei ∈
C(G). By Cauchy’s Theorem there exist x1 , . . . , xk ∈ G and H1 , H2 , . . . , Hk < G,
where Hi = hxi i and |Hi | = pi . Since G has only one nontrivial cyclic subgroup,
it follows that G has only one prime factor, hence |G| = pα for some prime p and
α ∈ Z+ .
Now, the only nontrivial cyclic subgroup of G is of order p. By Lagrange,
the possible orders of elements in G are 1, p, . . . , pα−1 , pα . If there exists an element
of order pβ , β ∈ {2, . . . , α}, then |C(G)| > 2 by Corollary 2.7, contrary to hypothesis. Then every nonidentity element of G has order p and must generate the same
cyclic subgroup by hypothesis. Thus, α = 1, so that |G| = p, or G = Cp for some
prime p.
This completes the proof.
7
Observe that the only groups in Table 2 satisfying |C(G)| = 3 are cyclic
groups of order p2 for p = 2, 3. This fact holds for any prime p as we show below.
The proof relies upon a corollary of the fact that any group of prime power order
has a non-trivial center, which we state as a lemma without proof. It also relies on
knowing |C(Cp × Cp )|; we classify |C(G)| for when G is the elementary abelian pgroup Cpn .
LEMMA 3.3: If |G| = p2 for some prime p, then G is abelian. More precisely, G is
isomorphic to either Cp2 or Cp × Cp . [2]
LEMMA 3.4: Let p be prime. If G = Cpn , then |C(G)| = 2 +
n−1
P
pk .
k=1
Proof. Let G = Cp × Cp · · · × Cp . Since G is an elementary abelian p-group, every
nonidentity element has order p. Then G contains pn − 1 elements of order p. Each
resulting cyclic subgroup has order p and can therefore be generated by any of the
p − 1 non-identity elements in the subgroup. Thus, the number of cyclic subgroups
of order p in G is given by
(p − 1)(pn−1 + · · · + p + 1)
pn − 1
=
= 1 + p + · · · + pn−1 .
p−1
p−1
Since the above sum does not include the trivial subgroup, it follows that
n−1
P k
p .
|C(G)| = 2 + p + · · · + pn−1 = 2 +
k=1
THEOREM 3.5: Let G be a finite group. Then |C(G)| = 3 if and only if G = Cp2
for some prime p.
Proof. (⇐) Suppose that G = Cp2 for some prime p. Then G is cyclic and has a
unique cyclic subgroup for every divisor of p2 , and since τ (p2 ) = 3, it follows that
|C(G)| = 3 by Proposition 2.3.
(⇒) Suppose that |C(G)| = 3 and write |G| = pα1 1 pα2 2 · · · pαk k . By Cauchy’s
Theorem there exist x1 , x2 , . . . , xk ∈ G and H1 , H2 , . . . , Hk < G, where Hi = hxi i
8
and |Hi | = pi . Since each Hi is of prime order, each Hi is cyclic, and by hypothesis it follows that G has at most two prime factors, say p1 and p2 , and at most two
corresponding cyclic subgroups H1 and H2 .
Case 1: |G| has one prime factor.
If |G| has only one prime factor, then |G| = pα for some α ∈ Z+ . If α = 1,
then |C(G)| = 2 by Proposition 3.2. If α = 2, then by Lemma 3.3 either G = Cp2 or
G = Cp × Cp . If G = Cp2 then |C(G)| = 3 by above. If G = Cp × Cp then by Lemma
3.4 |C(G)| = p + 2, and p + 2 = 3 ⇐⇒ p = 1, a contradiction.
Suppose that α ≥ 3. By Cauchy there exists x ∈ G, |x| = p. There cannot
exist an element of order pβ , β ≥ 3 and β ≤ α, since then |C(G)| ≥ 4. Thus every
nonidentity element of G has order p or p2 . By hypothesis G has one remaining
cyclic subgroup, say H, which then must have order p or p2 .
Consider |H| = p. Every nonidentity element not in hxi must be in H. Now
|G| − |hxi| = pα − p = p(pα−1 − 1).
Thus |G| − |hxi| = |H| − 1, or p(pα−1 − 1) = p − 1, which implies that p | (p − 1), a
contradiction.
Now consider |H| = p2 . Then by Cauchy there exists y ∈ H, |y| = p. If
x 6= y, then |C(G)| = 4, a contradiction. If x = y, then hxi < H, so that G contains
pα − p2 = p2 (pα−2 − 1) nonidentity elements not in H. There must always be such
elements, since |G| − |H| = 0 ⇐⇒ pα − p2 = 0 ⇐⇒ pα = p2 ⇐⇒ α = 2, but
α ≥ 3. Each of these elements generates a cyclic subgroup not contained in H. But
H contains all the cyclic subgroups of G, a contradiction.
Case 2 |G| has two prime factors.
Write |G| = pα1 1 pα2 2 . By Cauchy there exist x, y ∈ G such that |hxi| = p1 and
|hyi| = p2 . Thus, |C(G)| ≥ 3. We show that |C(G)| > 3. Consider hxyi. Without
9
loss of generality suppose that xy ∈ hxi. Then xy = xk for some k ∈ Z+ . By
cancellation y = xk−1 , so that e = y p2 = (xk−1 )p2 , hence |xk−1 | | p2 . But |xk−1 | = p1 ,
so that p1 | p2 , a contradiction. Thus hxyi 6∈ hxi, so that hxyi 6< hxi. Similarly,
hxyi 6< hyi, and therefore |C(G)| ≥ 4.
In either case |C(G)| = 3 follows only when G = Cp2 .
This completes the proof.
The case when |C(G)| = 4 is considerably more involved than the previous
three results and relies upon two lemmas.
LEMMA 3.6: Let G be a group of order pq, where p and q are prime with p < q.
There is only one subgroup of G of order q. [3]
LEMMA 3.7: All of the Sylow subgroups of a finite group are normal if and only
if the group is isomorphic to the direct product of its Sylow subgroups. [4]
We are now ready to characterize which groups satisfy |C(G)| = 4.
THEOREM 3.8: Let G be a finite group. Then |C(G)| = 4 if and only if G =
C2 × C2 , G = Cpq , or G = Cp3 .
Proof. (⇐) By Lemma 3.4 |C(C2 × C2 )| = 4, and by Proposition 2.3 |C(Cpq )| =
τ (pq) = 4 and |C(Cp3 )| = τ (p3 ) = 4.
(⇒) Assume |C(G)| = 4 and write n = |G| = pα1 1 pα2 2 · · · pαk k . By Cauchy’s
Theorem and the hypothesis we must have k ≤ 3, so |G| = pα1 1 pα2 2 pα3 3 . We consider
three cases.
Case 1: |G| = pα (|G| has one prime factor)
If α = 1, then G = Cp and |C(G)| = 2 by Proposition 3.2, a contradiction.
If α = 2, then |G| = p2 and by Lemma 3.3 G is abelian. Moreover G ∼
= Cp2
or G ∼
= Cp × Cp . If G = Cp2 , then |C(G)| = 3 by Theorem 3.5, a contradiction. If
G = Cp × Cp , then by Lemma 3.4 |C(G)| = p + 2, hence |C(G)| = 4 ⇐⇒ p = 2, so
that G = C2 × C2 .
10
If α = 3, then |G| = p3 . If G is abelian, then G is isomorphic to one of Cp3 ,
Cp × Cp2 , or Cp × Cp × Cp . If G = Cp × Cp × Cp , then |C(G)| = 2 + p + p2 by Lemma
3.4, hence |C(G)| = 4 ⇐⇒ p+p2 = 2 ⇐⇒ p = 1, a contradiction. If G = Cp ×Cp2 ,
write G = hxi × hyi, where |x| = p and |y| = p2 . Since hyi = Cp2 , hyi contains an
element of order p, say z. Then G has cyclic subgroups hei, Cp × hei = h(x, e)i,
hei × Cp = h(e, z)i, Cp × Cp = h(x, z)i, and hei × Cp2 = h(e, y)i, hence |C(G)| ≥ 5, a
contradiction. If G = Cp3 , then |C(G)| = 4.
Now suppose that G is non-abelian. For p = 2, G = D8 or G = Q8 . By
Table 2, |C(D8 )| = 7 and |C(Q8 | = 5. For an odd prime p, by p. 179 of Dummit
and Foote [2] there are precisely two non-abelian groups of order p3 (up to isomorphism). The first is the Heisenberg Group over Cp , which has the presentation
hx, a, b | xp = ap = bp = e, ab = ba, xax−1 = ab, xbx−1 = bi.
Since x, a, b ∈ G are distinct elements of order p it follows immediately that |C(G)| ≥
4. Consider ab ∈ G. If ab ∈ hai or ab ∈ hbi, then ab = ak or ab = bk for some
k ∈ Z+ , hence b = ak−1 or a = bk−1 , a contradiction. Suppose then that ab = xk for
some k ∈ Z+ . Then by the presentation xk = ab ⇐⇒ xk = xax−1 ⇐⇒ xk = a,
a contradiction. Since a and b commute it follows that habi is a distinct cyclic subgroup of order p, hence |C(G)| ≥ 5, a contradiction.
The other non-abelian group G of order p3 has the presentation
2
hx, y | xp = y p = e, xyx−1 = y 1+p i.
Let z = y p . Then G has the cyclic subgroups hei, hxi, hzi, and hyi, hence |C(G)| ≥
4. Consider xy ∈ G. Then xy = xk ⇐⇒ y = xk−1 , a contradiction, hence xy 6∈ hxi.
Similarly, for k ∈ {1, . . . , p − 1}, xy = z k =⇒ x = y pk−1 ∈ hyi, a contradiction.
Also xy = y k =⇒ x = y k−1 ∈ hyi, a contradiction. Thus hxyi is a distinct cyclic
11
subgroup and |C(G)| ≥ 5.
Now consider |G| = pα for α ≥ 4. By Cauchy there exists x ∈ G, |x| = p. By
hypothesis G cannot contain an element of order pβ , β ≥ 4 and β ≤ α, since then
|C(G)| ≥ 5. Thus every nonidentity element has order p, p2 , or p3 .
Suppose that G contains an element of order p3 , say y. Then hyi = Cp3 < G
and |C(hyi)| = 4, so that by hypothesis hyi contains all the cyclic subgroups of G.
There are then pα − p3 = p3 (pα−3 − 1) (nonidentity) elements in G not contained in
hyi. Each of these elements generates a cyclic subgroup not contained in hyi. But
hyi contains all of the cyclic subgroups of G, a contradiction. Thus every nonidentity element of G has order p or p2 . The two remaining cyclic subgroups of G, say
H and K, must then have orders of either p or p2 .
Subcase 1: |H| = |K| = p.
In this case the only nontrivial cyclic subgroups of G have order p. By hypothesis H, K, and hxi must intersect trivially (otherwise |C(G)| < 4), and every
nonidentity element of G is contained in one of these cyclic subgroups. Thus, every
nonidentity element of G has order p, so that G is the elementary abelian p-group
Cpα . Since α ≥ 4, by Lemma 3.4
|C(G)| = 2 +
α−1
X
pk = 2 + p + p2 + p3 ≥ 16,
k=1
which is a contradiction.
Subcase 2: |H| = p, |K| = p2 (Without loss of generality).
By Cauchy there exists z ∈ K, |z| = p. If hzi 6= hxi or hzi 6= H, then
|C(G)| = 5, a contradiction. Thus hzi must be one of hxi or H. Without loss of
generality suppose that hzi = H. Then H < K. There are then |G| − |K| − |hxi| +
1 nonidentity elements in G not contained in K or hxi. Moreover, G must always
12
have such “leftover” elements, since
|G| − |K| − |hxi| + 1 = 0 ⇐⇒ pα − p2 − p + 1 = 0
⇐⇒ pα − p2 = p − 1 ⇐⇒ p(pα−1 − p) = p − 1
⇐⇒ p | (p − 1)
a contradiction. Each of these elements generates a cyclic subgroup not contained
in K or hxi. But the cyclic subgroups of G are precisely those contained in K or
hxi. This is a contradiction.
Subcase 3: |H| = |K| = p2
By hypothesis we must have that H 6= K. By Cauchy there exists h ∈ H,
k ∈ K such that |h| = |k| = p. If one of hhi or hki equals hxi then |C(G)| = 5.
If neither hhi nor hki equals hxi then |C(G)| = 6. Therefore we must have that
hhi = hki = hxi. It follows that |H ∩K| = p. There are then |G|−|H|−|K|+|H ∩K|
nonidentity elements in G not contained in H or K. There must always be such
“leftover” elements, since
|G| − |H| − |K| + |H ∩ K| = 0 ⇐⇒ pα − p2 − p2 + p = 0
⇐⇒ pα − p2 = p2 − p ⇐⇒ p2 (pα−2 − 1) = p(p − 1) ⇐⇒
p(pα−2 − 1) = (p − 1) ⇐⇒ p | (p − 1)
a contradiction. Each of these elements generates a cyclic subgroup not contained
in H or K. But the cyclic subgroups of G are precisely those contained in H or K.
This is a contradiction.
To summarize Case 1, |C(G)| = 4 implies that G = Cp3 or G = C2 × C2 .
13
Case 2: |G| = pα q β (|G| has two prime factors)
Without loss of generality suppose that p < q. By Cauchy let x, y ∈ G such
that |x| = p, |y| = q. Observe that hxyi is a distinct cyclic subgroup from hxi or
hyi. Thus G has four distinct cyclic subgroups: hxi, hyi, hxyi, and hei. By Sylow 1
there exist Sylow subgroups A, B < G such that |A| = pα and |B| = q β . Let a ∈ A
and consider hai. Then |a| must divide pα , and since |C(G)| = 4 we must have that
hai = hxi. Since a ∈ A is arbitrary it follows that A = hxi. By a similar argument
B = hyi, hence α = β = 1 and |G| = pq.
Either G is abelian or non-abelian. If G is abelian then by page 143 of Dummit and Foote [2] G is cyclic, so that G = Cpq and |C(G)| = τ (pq) = 4.
If G is non-abelian, let x, y ∈ G, where |x| = p, |y| = q. By Lemma 3.6 hyi is
the unique cyclic subgroup of order q in G. Thus, |C(G)| ≥ 3. Now, the cyclic subgroups hxi and hyi account for p − 1 and q − 1 distinct elements in G, respectively.
Including the identity element this accounts for (p − 1) + (q − 1) + 1 = p + q − 1
distinct elements in G. There are then
pq − (p + q − 1) = pq − p − q + 1 = p(q − 1) − (q − 1) = (p − 1)(q − 1)
remaining elements in G which must be of order p. Each of these elements generates a cyclic subgroup which accounts for p − 1 distinct elements in the group. Thus
there are q − 1 of these elements in G.
When q = 2, there exists 1 distinct cyclic subgroup of order p not equal to
hxi, hence |C(G)| = 4, but q = 2 implies that p < 2, a contradiction.
When q = 3, there are 2 distinct cyclic subgroups of order p in G not equal
to hxi, hence |C(G)| = 5, a contradiction.
For any prime q > 3, q − 1 > 2, hence |C(G)| > 5, a contradiction.
Thus in this case |C(G)| = 4 only when G = Cpq .
14
Case 3: |G| = pα q β rγ (|G| has three prime factors)
By Cauchy’s Theorem again there exists x, y, z ∈ G such that |x| = p,
|y| = q, |z| = r, and by hypothesis G does not contain any more cyclic subgroups (excluding the trivial subgroup). By Sylow 1 there exist Sylow subgroups
A, B, C < G with |A| = pα , |B| = q β , |C| = rγ . Let a ∈ A and consider hai.
Then |a| must divide pα , so that hai = hxi. Thus A = hxi, and by similar arguments we obtain B = hyi and C = hzi. Hence |G| = pqr. Now either all the Sylow
subgroups are normal in G or at least one of the Sylow subgroups A, B, C is not
normal in G. In the first case, by Lemma 3.7 G is isomorphic to the direct product
of A, B, and C, hence G ∼
= A × B × C = Cp × Cq × Cr . Then G contains isomorphic copies of Cp , Cq , and Cr , hence |C(G)| = 4, but G also contains an isomorphic
copy of Cp × Cq ∼
= Cpq , hence |C(G)| ≥ 5, a contradiction. For the second case,
suppose without loss of generality that A 6E G. Then np ≥ 2, so that G contains a
distinct Sylow p-subgroup from A, say D, where |D| = p. Then D is cyclic, hence
|C(G)| ≥ 5, a contradiction.
Thus |C(G)| 6= 4 when |G| has three distinct prime factors.
This completes the proof.
We conclude the first section with a classification of |C(G)| for the dihedral
group G = D2n .
PROPOSITION 3.9: For all n ∈ N, |C(D2n )| = n + τ (n).
Proof. Let n ∈ N, and consider
D2n = hr, s | rn = s2 = e, rs = sr−1 i.
By the presentation of D2n , hsi is a cyclic subgroup of order 2. Observe that for
each k ∈ {1, 2, . . . , n − 1}, hsrk i is a distinct cyclic subgroup of order 2. For, given
15
k,
(srk )2 = srk srk = srk r−k s = sr0 s = s2 = e.
Since each of the srk are distinct, it follows that there are n − 1 cyclic subgroups
of order 2 generated by the srk . Consider now the rotational subgroup hri. Since
hri ∼
= Cn , it follows that the number of cyclic subgroups of hri is equal to the number of cyclic subgroups of Cn . Then by Corollary 2.7 |C(hri)| = τ (n), so that
|C(D2n )| = 1 + (n − 1) + τ (n)
= n + τ (n)
which proves the claim.
16
4. HIGHER END RESULTS FOR |C(G)|.
In this section we turn our attention to classifying which finite groups have
a large number of cyclic subgroups relative to their order. M. Tărnăuceanu states a
result for |C(G)| = |G| in his paper, which we prove. We also supply an alternate
proof of his result that characterizes the finite groups satisfying |C(G)| = |G| −
1. Finally, we expand the open problem in the paper by characterizing the finite
groups satisfying |C(G)| = |G| − 2.
We begin by considering the case when the order of G and the number of
distinct cyclic subgroups of G are equal. In this case every distinct element generates a distinct cyclic subgroup in G. By Table 1 it appears that the groups of small
order satisfying |C(G)| = |G| are the trivial group and the elementary abelian 2groups G = C2n for n = 1, 2, 3. In fact, the only finite groups satisfying |C(G)| = |G|
are precisely G = hei and those in this infinite family, as we now prove.
THEOREM 4.1: Let G be a finite group. Then |C(G)| = |G| if and only if G =
C2n for n ∈ N ∪ {0}.
Proof. (⇐) The result is obvious when G = hei. Let G = C2n for some n ∈ N. Then
every nonidentity element of G has order 2, so that any two distinct elements in G
generate distinct cyclic subgroups, hence |C(G)| = |G|.
(⇒) Suppose that |G| = |C(G)|. Then the map ψ : G → C(G) defined by
ψ(x) = hxi is a bijection. Let x ∈ G and suppose that m = |x| > 2. For every
positive integer m, (m − 1, m) = 1. Set k = m − 1 and y = xk = x−1 . Observe
that x 6= y since |x| = m > 2. Then by Proposition 2.1 |y| =
m
(k,m)
= m, so that y
also generates hxi. But then ψ(x) = ψ(y) and x 6= y, so that ψ is not a bijection, a
contradiction. Thus m ∈ {1, 2}. If G contains no elements of order 2 then G is the
trivial group. Otherwise, every nonidentity element of G has order 2.
17
Thus G = C2n , so that G is the elementary abelian 2-group.
It is simple to show that Theorem 4.1 is the maximal case in the relationship of |C(G)| and |G| for finite groups.
PROPOSITION 4.2: For every finite group G, |C(G)| ≤ |G|.
Proof. Let G be a finite group and suppose that |C(G)| > |G|. Then the map φ :
G → C(G) defined by φ(x) = hxi is not onto, so that there exists H ∈ C(G) such
that φ(g) 6= H for all g ∈ G. But H must be generated by some element in G. This
is a contradiction.
We base our proof of the next two results on Tărnăuceanu’s proof for the
|C(G)| = |G| − 1 case. In order to do so, we isolate Tărnăuceanu’s “set-up” of the
problem as a remark.
REMARK 4.3: Let |G| = n and d1 = 1, d2 , . . . , dk be the positive divisors of n. If
P
ni = |{H ∈ C(G) | |H| = di }|, i ∈ {1, 2, . . . , k}, then ki=1 ni φ(di ) = n, where φ(x)
is Euler’s phi function. [1]
The identity above holds since every element x ∈ G generates a cyclic subgroup whose order di is one of the divisors of n by Proposition 2.3. Each cyclic
subgroup has φ(di ) generators by Proposition 2.2. Thus the total number of elements in G is equal to the sum of the number of distinct cyclic subgroups of order
di (i = 1, . . . , k) times the number of generators of each subgroup.
Our proof also relies on three elementary results. The first result states that
the order of an element in a finite group is invariant under conjugation. The second and third result, taken together, characterize when the product set HK of two
subgroups H, K < G is itself a subgroup of G.
PROPOSITION 4.4: Let G be a finite group and x ∈ G with |x| = n. Then
|gxg −1 | = n ∀ g ∈ G.
18
Proof.
(gxg −1 )n = gxg −1 gxg −1 . . . gxg −1
= gx(g −1 g)x(g −1 g)x(g −1 g) . . . (g −1 g)xg −1
= gxn g −1 = gg −1 = e
If k < n, then the calculation of (gxg −1 )k upon cancellation will yield gxk g −1 , so
that the minimality of |x| = n forces |gxg −1 | = n.
PROPOSITION 4.5: If H and K are subgroups of a group G, then HK < G if
and only if HK = KH. [2]
COROLLARY 4.6: If H, K < G and one of H or K are normal in G then HK <
G.
Proof. (Corollary 4.6) Without loss of generality consider H E G. Then gH = Hg
for all g ∈ G, so that upon restriction to elements in K, kH = Hk for all k ∈ K, or
KH = HK.
We now present an alternate proof characterizing the finite groups G satisfying |C(G)| = |G| − 1. Our proof follows Tărnăuceanu’s own up until the bullet
points.
THEOREM 4.7: Let G be a finite group. Then |C(G)| = |G| − 1 if and only if G
is one of C3 , C4 , S3 , or D8 .
Proof. (⇐) Observe that |C(C3 )| = τ (3) = 2 and |C(C4 )| = τ (4) = 3 by Proposition
2.3, and |C(S3 )| = |C(D6 )| = 3+τ (3) = 5 and |C(D8 )| = 4+τ (4) = 7 by Proposition
3.9, so that the result holds for each group.
(⇒) Let G be a finite group of order n and suppose that |C(G)| = |G| − 1.
P
By Remark 4.3 we have ki=1 ni φ(di ) = n, where d1 = 1, . . . , dk are the positive
19
divisors of n and ni = |{H ∈ C(G) | |H| = di }|. By hypothesis we have that
P
|C(G)| = ki=1 ni = n − 1, hence
k
X
ni (φ(di ) − 1) = 1.
i=1
This implies that
• There exists i0 ∈ {1, 2, . . . , k} such that ni0 = 1 and φ(di0 ) = 2 (i.e. di0 ∈
{3, 4, 6});
• For i 6= i0 either ni = 0 or φ(di ) = 1 (i.e. di ∈ {1, 2}).
Thus G has a unique cyclic subgroup of order 3, 4, or 6, and arbitrarily (finitely)
many cyclic subgroups of order 2. Moreover, |G| = 2α 3β . Observe that G cannot contain an element of order 6, for then G also contains an element of order 3,
a contradiction. We consider two cases and prove each by a series of claims.
Case 1: G has a unique cyclic subgroup of order 3.
Let H = hhi, |h| = 3.
Claim 1: H E G.
We show gHg −1 ⊆ H ∀ g ∈ G. Certainly geg −1 ∈ H ∀ g ∈ G. Let x ∈ H,
x 6= e. Then |gxg −1 | = 3 ∀ g ∈ G, and since H is the unique cyclic subgroup
of order 3 in G, H must contain all elements of order 3 in G. Thus gxg −1 ∈ H
∀ g ∈ G, so that gHg −1 ⊆ H ∀ g ∈ G and H E G.
Claim 2: H is a Sylow 3-subgroup of G.
By Sylow 1 there exists A < G, |A| = 3β . By hypothesis A cannot contain
an element of order 3β for β ≥ 2. Thus every nonidentity element a ∈ A must have
order 3. But H contains all elements of G of order 3, so that we must have A = H.
This proves the claim. Moreover, β = 1, so that |G| = 2α 3.
Now either H = G or H < G. If H = G then G = C3 . Suppose then that H < G.
20
Claim 3: ghg −1 = h2 ∀ g ∈ G − H.
Let g ∈ G − H. Then |ghg −1 | = 3, so that ghg −1 = h or ghg −1 = h2 . If
ghg −1 = h, then gh = hg ∀ g ∈ G. By hypothesis we must have |g| = 2. Since
g and h commute it follows that |gh| = lcm(|g|, |h|) = 6, a contradiction since G
contains no elements of order 6. Thus ghg −1 = h2 , or ghg −1 = h−1 .
Now let k ∈ G − H and K = hki. By hypothesis |K| = 2, and since k ∈
/ H,
H ∪ K = {e}. Moreover, since H E G it follows that HK < G by Corollary 4.6.
Claim 4: G = HK
Suppose that HK 6= G. Then ∃ x ∈ G, x ∈
/ HK, and by hypothesis |x| = 2.
Consider conjugation by xk. By Claim 3 (xk)h(xk)−1 = h2 ⇐⇒ xkhk −1 x−1 = h2 .
By Claim 3 again khk −1 = h2 , hence xh2 x−1 = h2 ⇐⇒ xh2 = h2 x. Thus, x and h2
commute, so that |xh2 | = lcm(|x|, |h2 |) = 6, a contradiction.
Thus G = HK, so that |G| = 6 and G has the presentation
hh, k | h3 = k 2 = e, kh = h−1 ki
whence it follows that G ∼
= S3 .
= D6 ∼
Case 2: G has a unique cyclic subgroup of order 4.
Let H = hhi, where |h| = 4. Observe that by hypothesis we must have
β = 0, so that |G| = 2α .
Claim 1: H E G.
Certainly geg −1 ∈ H ∀ g ∈ G. Let x ∈ H, x 6= e, so that |x| = 2 or |x| = 4.
If |x| = 4, then x = h or x = h3 , and |gxg −1 | = 4 ∀ g ∈ G. Since H contains all
the elements of order 4 in G by uniqueness it follows that gxg −1 ∈ H ∀ g ∈ G. If
|x| = 2, then x = h2 and |gxg −1 | = 2 ∀ g ∈ G. Suppose that gxg −1 ∈
/ H for some
g ∈ G. Then g ∈
/ H. Set y = gxg −1 . Then
gxg −1 = y ⇐⇒ gh2 g −1 = y ⇐⇒ ghg −1 ghg −1 = y ⇐⇒ (ghg −1 )2 = y.
21
By the previous case ghg −1 ∈ H and |ghg −1 | = 4, hence ghg −1 = h or ghg −1 = h3 .
In either case (ghg −1 )2 = h2 , hence y = h2 ∈ H, a contradiction. Thus gxg −1 ∈ G
for all g ∈ G, x ∈ H, so that H E G.
Now either H = G or H < G. If H = G then G = C4 . Suppose then that H < G.
Claim 2: ghg −1 = h−1 ∀ g ∈ G − H.
Let g ∈ G − H. Then |g| = 2 and |ghg −1 | = 4, and since H E G either
ghg −1 = h or ghg −1 = h3 = h−1 . If ghg −1 = h, then gh = hg. Since g and h
commute, |gh| = lcm(|g|, |h|) = 4, hence gh = h or gh = h3 . Then either g = e
or g = h2 , which both imply that g ∈ H, a contradiction. Thus ghg −1 = h−1 ∀ g ∈
G − H.
Now let k ∈ G − H and K = hki. Then |K| = 2 and since k ∈
/ H, H ∩ K =
{e}. Moreover, since H E G, HK < G.
Claim 3: G = HK.
Suppose that G 6= HK. Then there exists x ∈ G, x ∈
/ HK. By hypothesis
|x| = 2. Consider conjugation by xk. Then by Claim 2
(xk)h(xk)−1 = h−1 ⇐⇒ x(khk −1 )x−1 = h−1 ⇐⇒ xh−1 x−1 = h−1 .
By Claim 2 again xh−1 x−1 = h, so that the above holds if and only if h = h−1 ,
which implies that |h| = 2, a contradiction. Therefore G = HK, so that |G| = 8
and G has the presentation
hh, k | h4 = k 2 = e, kh = h−1 ki
hence G ∼
= D8 . This completes the proof.
We conclude this section by presenting a characterization of the finite groups
G satisfying |C(G)| = |G| − 2. The proof for this result is more involved than the
proof for the |C(G)| = |G| − 1 result as the number of cases to consider is effectively
22
tripled by the consequences of Remark 4.3.
In two cases of the proof we form internal semidirect products out of product sets of subgroups of G and then show that these semi-direct products are isomorphic to direct products. We make use of the following results concerning semidirect products:
PROPOSITION 4.8: Suppose G is a group with subgroups H and K such that
(1) H E G
(2) H ∩ K = {e}
Let φ : K → Aut(H) be the homomorphism defined by mapping k ∈ K to the
automorphism of left conjugation by k on H. Then HK ∼
= H o K. In particular, if
G = HK with H and K satisfying (1) and (2), then G is the semidirect product of
H and K. [2]
PROPOSITION 4.9: Let H and K be groups and let φ : K → Aut(H) be a
homomorphism. Then the following are equivalent:
(1) The identity (set) map between H o K and H × K is a group homomorphism
(hence an isomorphism)
(2) φ is the trivial homomorphism from K into Aut(H)
(3) K E H o K. [2]
THEOREM 4.10: Let G be a finite group. Then |C(G)| = |G| − 2 if and only if G
is one of the following groups: C4 × C2 , C2 × D8 , C6 , or D12 .
Proof. (⇐) By Proposition 2.3 |C(C6 )| = τ (6) = 4, and by Proposition 3.9 |C(D12 )| =
6 + τ (6) = 10. That the result holds for the other two groups follows by hand calculation.
23
(⇒) Suppose that |C(G)| = |G| − 2 and let |G| = n. Then by Remark 4.3
Pk
n =
i=1 ni φ(di ), where d1 = 1, . . . , dk are the positive divisors of n and ni =
P
|{H ∈ C(G) | |H| = di }|. By hypothesis we have that |C(G)| = ki=1 ni = n − 2,
hence
k
X
ni (φ(di ) − 1) = 2
i=1
which implies that one of the following must hold:
a.) There exists i0 ∈ {1, 2, . . . , k} such that ni0 = 1 and φ(di0 ) = 3.
b.) There exists i0 ∈ {1, 2, . . . , k} such that ni0 = 2 and φ(di0 ) = 2 (i.e. di0 ∈
{3, 4, 6}) and
i.) For an i 6= i0 either ni = 0 or φ(di ) = 1 (i.e. di ∈ {1, 2}).
c.) There exist i, j ∈ {1, . . . , k} such that ni = nj = 1 and φ(di ) = φ(dj ) = 2
(hence di , dj ∈ {3, 4, 6}) and
i.) For l 6= i and l 6= j either nl = 0 or φ(dl ) = 1 (i.e. dl ∈ {1, 2}).
Since φ(di0 ) 6= 3 for all di0 ∈ N, we may discard a.). In the remaining cases G has
two cyclic subgroups H and K of orders 3, 4, or 6. Under b.), we obtain |H| = |K|,
so that b.) yields the three subcases: (|H|, |K|) = (3, 3), (|H|, |K|) = (4, 4), and
(|H|, |K|) = (6, 6). Under c.), we obtain that |H| 6= |K|, so that c.) yields the three
subcases: (|H|, |K|) = (3, 4), (|H|, |K|) = (3, 6), and (|H|, |K|) = (4, 6). In either
b.) or c.) G contains arbitrarily (finitely) many elements of order 2, and no other
cyclic subgroups. Since the subcases will themselves contain subcases we simply
write the subcases as cases 1 through 6 with the understanding that the first three
cases proceed from b.) and that the last three proceed from c.).
Case 1: (|H|, |K|) = (3, 3).
Let H = hxi, K = hyi, where x 6= y. Consider xyx−1 ∈ G. Then |xyx−1 | = 3.
Suppose that xyx−1 ∈ H. Then xyx−1 = xk for some k ∈ Z+ , which implies that
24
y = xk ∈ H, a contradiction. Thus xyx−1 ∈ K. Then either xyx−1 = y or xyx−1 =
y2.
Subcase 1: xyx−1 = y.
If xyx−1 = y, then xy = yx. Since x and y commute, |xy| = 3, hence xy ∈ H
or xy ∈ K. If xy ∈ H, then xy = xk for some k ∈ Z+ , hence y = xk−1 ∈ H, a
contradiction. Similarly, if xy ∈ K, we obtain x = y k−1 ∈ K, a contradiction.
Subcase 2: xyx−1 = y 2 .
In this subcase xy = y 2 x. Consider |xy|.
xy = y 2 x
(xy)2 = xyxy = xyy 2 x = x2
(xy)3 = (xy)2 xy = x2 xy = y
(xy)4 = (xy)2 (xy)2 = x4 = x
(xy)5 = (xy)(xy)4 = (y 2 x)x = y 2 x2
(xy)6 = (xy)2 (xy)4 = x3 = e.
Thus |xy| = 6, so that G contains a cyclic subgroup of order 6, a contradiction.
Thus, Case 1 does not hold.
Case 2: (|H|, |K|) = (4, 4).
Let x, y ∈ G such that x 6= y and H = hxi and K = hyi. Consider yxy −1 .
Then |yxy −1 | = 4, so either yxy −1 ∈ H or yxy −1 ∈ K. If yxy −1 ∈ K, yxy −1 = y k
for some k ∈ Z+ , hence x = y k−1 ∈ K, a contradiction. Thus, yxy −1 ∈ H, so that
either yxy −1 = x or yxy −1 = x−1 .
Subcase 1: yxy −1 = x−1 .
25
In this subcase yx = x−1 y. Consider |yx|. Then
yx = x−1 y
(yx)2 = yxyx = yxx−1 y = y 2
(yx)3 = yx(yx)2 = yxy 2
(yx)4 = yx2 (yx)2 = y 2 y 2 = e.
Thus |yx| = 4, so that yx ∈ H or yx ∈ K. If yx ∈ H, then either yx = x or
yx = x3 , which implies that either y = e ∈ H or y = x2 ∈ H, a contradiction.
Similarly, yx ∈ K implies that either x = e ∈ K or x = y 2 ∈ K, a contradiction.
Thus yxy −1 6= x−1 .
Subcase 2: yxy −1 = x.
In this case yx = xy. Either H ∩ K = {e} or H ∩ K 6= {e}.
Subcase 2a: H ∩ K = {e}.
Since the generators of H and K commute, HK < G. Then |HK| =
|H||K|
|H∩K|
=
16. Consider xy ∈ HK. Then |xy| = lcm(|x|, |y|) = 4, hence xy is one of x, x−1 , y,
or y −1 . These conditions imply that either y = e, y = x2 , x = e, or x = y 2 , respectively. Each is a contradiction.
Subcase 2b: H ∩ K 6= {e}.
In this case |H ∩ K| = 2 or |H ∩ K| = 4. If |H ∩ K| = 4, then H ∩ K = H
or H ∩ K = K, hence H = K, a contradiction. Thus |H ∩ K| = 2, so that H ∩ K =
{e, x2 } = {e, y 2 }, hence x2 = y 2 . Consider hx, yi = HK. Then |hx, yi| = |HK| = 8.
We obtain the presentation
HK = hx, y | x2 = y 2 , x4 = y 4 = e, xy = yxi.
26
The orders of the nonidentity elements in HK are given below:
• Elements of order 4: x, x2 , y, y = x2 y
• Elements of order 2: x2 = y 2 , xy, x3 y = xy 3
Thus HK is an abelian group containing four elements of order 4 and three elements of order 2. Up to isomorphism there are three abelian groups of order 8,
namely C8 , C23 , and C4 × C2 . Of these, C4 × C2 is the only group with the same
order structure as HK. It follows that HK ∼
= C 4 × C2 .
Now either G = HK or HK < G. If G = HK then G ∼
= C4 × C2 and we are
done. Suppose then that HK < G. Then there exists z ∈ G such that z ∈
/ HK and
|z| = 2. Let I = HK and J = hzi.
Claim 1: I E G.
We show that if g ∈ G, gIg −1 ⊆ I. Let a ∈ I and g ∈ G. Then a = xi y j for
some i, j ∈ {1, . . . , 4}. Observe that
gag −1 = gxi y j g −1
= gxg −1 gxi−1 g −1 gyg −1 gy j−1 g −1
= · · · = (gxg −1 )i (gyg −1 )j .
Thus it suffices to show that gxg −1 ∈ hxi = H and gyg −1 ∈ hyi = K for all g ∈ G.
Lemma: gxg −1 ∈ H for all g ∈ G.
Since |gxg −1 | = 4, gxg −1 ∈ {x, x−1 , y, y −1 }. If gxg −1 = x or gxg −1 = x−1 , we
are done. If gxg −1 = y, then gx = yg for all g ∈ G. In particular, when g = y, we
obtain yx = y 2 ⇒ x = y, a contradiction. If gxg −1 = y −1 , then gx = y −1 g for all
g ∈ G. In particular, when g = y, yx = e ⇒ y = x−1 ∈ H, a contradiction. Thus
gxg −1 ∈ H. By interchanging x and y above, we obtain gyg −1 ∈ K. Since g ∈ G is
27
arbitrary the lemma follows.
Thus, gag −1 = xα y β ∈ I, hence gIg −1 ⊆ I for all g ∈ G, so that I = HK E G.
Moreover, IJ < G by Corollary 4.6 and I ∩ J = {e}. Now let φ : J → Aut(I) be the
homomorphism defined by mapping z ∈ J to the automorphism of left conjugation
by z on I. Then by Proposition 4.8 IJ ∼
= I o J.
We now show that HJ = hx, zi ∼
= D8 . Observe that HJ < G since H E G
by the lemma.
Claim 2: zxz −1 = z −1 .
Consider zxz −1 ∈ G. Then |zxz −1 | = 4, hence zxz −1 ∈ {x, x−1 , y, y −1 }.
If zxz −1 = x, then zx = xz. Since z and x commute, it follows that |zx| =
lcm(|x|, |z|) = 4, so that xz ∈ {x, x−1 , y, y −1 }. The first two equations imply that
z ∈ H, and the second equations imply that z ∈ J = HK, a contradiction. If
zxz −1 = y, then zx = yz, and (zx)2 = zxzx = yzzx = yx, and since x and y commute zx must have order 4, but zx ∈
/ H and zx ∈
/ K (otherwise z ∈ H or z ∈ K). If
zxz −1 = y −1 , then zx = y −1 z. Now (zx)2 = zxzx = y −1 zzx = y −1 x, and since x and
y −1 commute zx must have order 4, but as before zx ∈
/ H and zx ∈
/ K.
Thus zxz −1 = x−1 , so that z conjugates x to its inverse, which proves the claim.
Observe that since z ∈ G is arbitrary the claim holds for any element a ∈ G
where a ∈
/ I = HK. Observe also that by interchanging x and y we obtain that any
element not in I also conjugates y to its inverse. The presentation for HJ is then
given by
HJ = hx, z | x4 = z 2 = e, zx = x−1 zi
hence it follows that HJ = hx, zi ∼
= D8 .
Claim 3: IJ = G.
Either IJ < G or IJ = G. Suppose that IJ < G. Then there exists a ∈ G
where a ∈
/ IJ, and by hypothesis |a| = 2. Now since a ∈
/ I, by Claim 2 and the
28
remark following the claim axa−1 = x−1 . Moreover, az ∈
/ I, hence
(az)x(az)−1 = azxz −1 a−1 = ax−1 a−1 = x
which implies that (az)x = x(az). Then, since az and x commute, it follows that
|(az)x| = lcm(|az|, |x|) = 4, but (az)x ∈
/ I, a contradiction. Thus G = IJ.
Claim 4: G = IJ ∼
= D8 × C2 .
Thus far we have established that G = IJ ∼
= I o J and that HJ = hx, zi ∼
=
D8 . Since [G : HJ] = 2, HJ E G. Let L = hxyi. Then |L| = 2 and HJ ∩ L = {e},
so that (HJ)L ∼
= HJ o L by Proposition 4.8. Since |(HJ)L| = 16, it follows that
G = (HJ)L ∼
= HJ o L ∼
= HJ × L ∼
=
= HJ o L ∼
= D8 o C2 . To show that D8 o C2 ∼
D8 × C2 , by Proposition 4.9 it suffices to show that L E HJ o L.
Claim 5: L E HJ o L.
Clearly geg −1 ∈ L for all g ∈ G. Consider xy ∈ L. Then, since every element
of G is of the form xi y j z k , by Claim 2
g(xy)g −1 = (xi y j z k )xy(xi y j z k )−1
= xi y j (z k xyz −k )y −j x−i
= xi y j x−1 y −1 y −j x−i
= x−1 y −1 = x3 y 3
= xx2 y 2 y = xy.
Thus g(xy)g −1 ∈ L for all g ∈ G, so that L E HJ o L. Finally, we obtain G ∼
=
D8 × C2 , which concludes the case.
Case 3: (|H|, |K|) = (6, 6).
If |H| = |K| = 6, then by Cauchy’s Theorem there exists h ∈ H, k ∈ K such
that |h| = |k| = 3, hence G contains two cyclic subgroups of order 3, contrary to
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hypothesis.
Case 4: (|H|, |K|) = (3, 4).
Let H = hxi and K = hyi. Consider yxy −1 ∈ G. Then |yxy −1 | = 3, hence
yxy −1 = x or yxy −1 = x2 . If yxy −1 = x, then yx = xy. Since x and y commute,
|xy| = lcm(|x|, |y|) = 12, so that G contains an element of order 12, a contradiction.
If yxy −1 = x2 , then yx = x2 y. Consider |yx|.
yx = x2 y
(yx)2 = (yx)(yx) = yxx2 y = y 2
(yx)3 = (yx)(yx)2 = yxy 2 = x2 yy 2 = x2 y 3
(yx)4 = (yx)(yx)3 = yxx2 y 3 = yx3 y 3 = y 4 = e
Thus |yx| = 4, so that yx ∈ K, hence yx = y or yx = y 3 . Then either x = e or
x = y 2 ∈ K, a contradiction.
Therefore Case 4.) does not hold.
Case 5: (|H|, |K|) = (6, 3).
Let H = hyi and hxi be the unique cyclic subgroup of order 3 in G. Then
hxi < H, for if it were not, then H contains a cyclic subgroup of order 3 not equal
to hxi, contradicting uniqueness. Now either H = G or H < G. If H = G, then
H ∼
/ H, and by
= C6 . Suppose then that H < G. Then there exists z ∈ G, z ∈
hypothesis |z| = 2. Set K = hzi.
Claim 1: zyz −1 = y −1
Consider zyz −1 ∈ G. Then |zyz −1 | = 6, hence zyz −1 = y or zyz −1 = y −1 . If
zyz −1 = y, then zy = yz and |yz| = lcm(|y|, |z|) = 6, so that yz = y or yz = y −1 ,
which implies that either z = e or z = y −2 ∈ H, a contradiction.
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Claim 2: H E G.
Since z ∈ G above is arbitrary, gyg −1 = y −1 ∈ H for all g ∈ G, where g ∈
/ H.
Clearly for all g ∈ H, gyg −1 ∈ H, so that H E G. It follows that HK < G by
Corollary 4.6. Since z ∈
/ H it is clear that H ∩K = {e}. Let φ : K → Aut(H) be the
automorphism defined by mapping k ∈ K to the automorphism of left conjugation
by k in H. Then HK ∼
= H o K < G by Proposition 4.8. By Claim 1 HK has the
presentation
HK = hy, z | y 6 = z 2 = e, zy = y −1 zi
so that HK ∼
= D12 .
Claim 3: G = HK
Suppose that there exists g ∈ G, g ∈
/ HK. Then by hypothesis |g| = 2. Also
zg ∈
/ HK, since otherwise g ∈ HK, and |zg| = 2. By Claim 1 it follows that
y −1 = zgy(zg)−1 = zgyg −1 z −1 = zy −1 z −1 = y
which implies that |y| = 2, a contradiction. Therefore G = HK, so that G ∼
= D12 .
Case 6: (|H|, |K|) = (4, 6)
If |K| = 6, then K contains a cyclic subgroup of order 3, but this contradicts
the fact that H is the only other cyclic subgroup not of order 2 in G.
This completes the proof.
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5. CONCLUSION
Classifying the number of distinct cyclic subgroups of a given finite group
G is an interesting problem that is surprisingly under-explored in the literature [1].
Moreover, the cyclic subgroup structure of a finite group is subject to some interesting constraints. Cauchy’s Theorem provides a natural bound for “lower end”
considerations of |C(G)|, while Tărnăuceanu’s identity in Remark 4.3 determines
what cases need to be considered when |C(G)| is large relative to |G|.
Our research establishes characterizations of |C(G)| = k for k ∈ {1, . . . , 4},
classifies |C(G)| for dihedral groups and elementary abelian p-groups, provides alternate proofs for |C(G)| = |G| − k, k ∈ {0, 1}, and also characterizes the case of
k = 2. The k = 2 case contributes to the open problem posed by Tărnăuceanu in
his article. Given the number of cases that arose for the k = 2 problem, it may be
necessary to find a stronger point of entry than Remark 4.3.
Another problem that we would like to explore in further research is to characterize when |C(G)| is a quotient of |G|. In particular, we are interested in obtaining classifications of finite groups satisfying |C(G)| =
|G|
k
for k ∈ N, k ≥ 2. We
supplied an original proof that the k = 1 case is satisfied only when G is an elementary abelian 2-group. We are interested to see if any patterns emerge among the
groups satisfying this relation as k increases.
We hoped to prove the following two conjectures but our other results took
one too many mornings to establish. We conclude our remarks by stating them
here.
CONJECTURE 5.1: Let G be a finite group. Then |C(G)| = 5 if and only if G is
isomorphic to one of the following groups: D6 , C3 × C3 , Q8 , or Cp4 .
CONJECTURE 5.2: Let G be a finite group. Then |C(G)| = |G| − 3 if and only
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if G is isomorphic to one of the following groups: C5 , Q8 , or D10 .
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REFERENCES
[1] M. Tărnăuceanu, Finite Groups With a Certain Number of Cyclic Subgroups,
Article, American Mathematical Monthly (AMM), 2015. (2016)
[2] Dummit, D., and Foote, F. Abstract Algebra, Wiley, Vermont, 2004. (2016)
[3] Conrad, Keith. Consequences of Cauchy’s Theorem, Lecture Notes: University
of Connecticut, math.uconn.edu/ kconrad/blurbs/grouptheory/cauchyapp.pdf.
(2016)
[4] Conrad, Keith. Consequences of the Sylow Theorems, Lecture Notes: University of Connecticut, math.uconn.edu/ kconrad/blurbs/grouptheory/sylowapp.pdf. (2016)
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